Thermal System filled

# Thermal System filled - ME375 Handouts Thermal Systems •...

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Unformatted text preview: ME375 Handouts Thermal Systems • Basic Modeling Elements – Resistance • Conduction • Convection • Radiation – Capacitance • Interconnection Relationships – Energy Balance - 1st Law of Thermodynamics • Derive Input/Output Models School of Mechanical Engineering Purdue University ME375 Thermal Systems - 1 Key Concepts • q : heat flow rate [J/sec = W] • T : temperature [oK] or [oC] ( ( ) ) Temperature in a body usually depends on spatial as well as temporal coordinates. As a temporal result, the dynamics of a thermal system has to be described by partial differential equations. Moreover, nonlinearities are often essential in describing the heat transfer by radiation and heat convection. However, very few nonlinear PDEs have analytical (closed form) solutions. Usually, finite element methods (FEM) are used to numerically solve nonlinear PDE solve problems. Our purpose is to try to use lumped model approximations of thermal systems to approximations obtain linear ODEs that are capable of describing the dynamic response of thermal systems to a good first approximation. For many thermal system, an equilibrium condition exists that defines the nominal operating defines condition. In these cases, the deviation of the heat flow rate and temperature from their nominal values, q and T , are of interest. Thus, we can define the incremental heat flow define rate ( q ( t ) ) and the incremental temperature ( T ( t ) ) to be: q (t ) = q (t ) − q and T (t ) = T (t ) − T School of Mechanical Engineering Purdue University ME375 Thermal Systems - 2 1 ME375 Handouts Basic Modeling Elements • Thermal Resistance Ex: Ex: Describes the heat transfer process through an element with the characteristic that the heat flow rate across the element is proportional to the temperature difference across the element, i.e. T1 q T2 + ΔT − T1 T2 q R Two bodies at temperatures T1 and T2 are separated by two elements with different thermal resistance R1 and R2. Heat flows through the two elements at a rate of q. Find the equivalent thermal resistance Req and solve for the interface temperature between the two elements. T1 R1 R2 T2 T1 R T2 Req q ΔT = T12 = T1 − T2 = R ⋅ q or 1 1 q = ( T1 − T2 ) = ΔT R R q [oK/W] R= School of Mechanical Engineering Purdue University ME375 Thermal Systems - 3 Three Types of Heat Transfer • Conduction Ex: Calculate the equivalent thermal Ex: resistance of a wall with a window. Heat transfer through solid or continuous Wall Window media via random molecular motion AG Area AW (diffusion). Thickness T1 q T2 d T α dW αW dG αG Cross sectional area A T1 T2 q= αA αA x (T1 − T2 ) = T12 ⇒ R = d d – α : thermal conductivity [W/m-oK] School of Mechanical Engineering Purdue University ME375 Thermal Systems - 4 2 ME375 Handouts Three Types of Heat Transfer • Convection Heat transfer between the interface of a solid material and a fluid material via bulk motion of the fluid. T TF q = hA ⋅ (TS − TF ) = hA ⋅ ΔT x q – A : surface area [m2] – h : convective heat transfer coefficient [W/m2-oK] – TS : surface temperature [oK] – TF : fluid temperature [oK] T TF TF x 1 hA – h depends on surface geometry, fluid flow rate, temperature, flow direction, ... ⇒ R= School of Mechanical Engineering Purdue University ME375 Thermal Systems - 5 Three Types of Heat Transfer • Radiation Except for radiation, both conductive Heat transfer via electromagnetic waves. and convective heat transfer processes T2 can be modeled as thermal T1 q resistances. In the previous discussions, the Surface Area A assumption is that the materials do not 4 4 store thermal energy. In reality, q = σ FE FV A ⋅ ( T1 − T2 ) materials do store a certain amount of – A : surface area [m2] thermal energy. – σ : Stefan-Boltzmann constant Stefan[W/m2-oK4] Q: How would we model the process of – FE : effective emissivity storing thermal energy ? – FV : view factor Nonlinear! Will not be considered in this course School of Mechanical Engineering Purdue University ME375 Thermal Systems - 6 3 ME375 Handouts Basic Modeling Elements • Thermal Capacitance The ability of a substance to hold or store heat is the heat capacity of the material and it behaves like a thermal capacitance. Since the specific heat cP can be interpreted as the heat storage capacity of the material per unit mass, the total heat storage capability of a material is: cP ⋅ M If there is net heat flow into the material, the temperature of the material will change and the rate of temperature change is proportional to the net heat flow rate qSTORE : d cP M TC = qSTORE = qIN − qOUT dt We can define the thermal capacitance C = cP M = cP ρ V + TC − TC qIN C qOUT qIN − qOUT C Mass, M Volume, V Density, ρ Note: The above relationship holds only if Note: we assume that the temperature is uniform across the entire material. School of Mechanical Engineering Purdue University ME375 Thermal Systems - 7 Interconnection Laws • Energy Balance - 1st Law of Thermodynamics – Energy stored in the system is the sum of the net energy inflow, the energy generated within the system and the work done on the system: q STORE = ∑q IN − ∑q OUT + q GENERATED ' + W ITHIN d WW ORK = dt DONE Ex: A material with a thermal capacitance C is surrounded by an insulation material with Ex: thermal resistance R. Heat is added to the inner material at a rate of qi(t). Find the system model, if the inner material temperature TC is to be the output. Ta TC , C R qi(t) School of Mechanical Engineering Purdue University ME375 Thermal Systems - 8 4 ME375 Handouts In Class Example Ex: The Pentium II processor under normal operation will generate heat at a rate of qi(t). The processor heat itself has a specific heat of cP. The cross sectional area of the chip is AP with a thickness of dP. The average density of the processor is ρP . To help dissipate the heat and reduce the processor temperature TP, a heat sink with the same cross sectional area and an average thickness of dS is added on top of the processor. The heat sink has a thermal conductivity of αS . To further improve heat conductivity dissipation, a fan is used to generate air flow on top of the heat sink, the effective convection heat coefficient is hA and the effective contact area between the heat sink and the air flow is AS . The air temperature inside the computer is maintained at TA. Find the relationship between qi(t) and the temperature of the processor TP. TA hA AP Heat Sink cP , ρP , TP School of Mechanical Engineering Purdue University dS dP ME375 Thermal Systems - 9 5 ...
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## This note was uploaded on 08/28/2010 for the course ME 375 taught by Professor Meckle during the Spring '10 term at Purdue.

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